On 30-1-2015 19:04, Mark Montague wrote:
On 2015-01-30 11:39, Abdul Abdul wrote:
Thanks for your reply. I get a blank screen.
On Fri, Jan 30, 2015 at 6:33 AM, Ashley Sheridan
<ash@xxxxxxxxxxxxxxxxxxxx>
wrote:
On 30 January 2015 14:26:51 GMT+00:00, Abdul Abdul
<abdul.sw84@xxxxxxxxx>
wrote:
I have the following PHP script, where I'm trying to retrieve the image
type. Why doesn't it work? What am I doing wrong?
<?php
$image="4.2.06"
$imagetype = exif_imagetype($image);
echo "Image type is: " $imagetype;
?>
What does "doesn't work" mean? What output are you getting? Are there
any
errors or warnings in your logs?
Ash's advice is correct: check your error log. However, I think that
one problem is that you don't have a semicolon at the end of the first
assignment statement above, which results in a syntax error occurring.
Also, please do not top-post. Put your replies at the bottom instead.
The problem is actually threefold;
1 - Errors aren't being displayed (display_errors in your php.ini is
turned OFF; turn it on to see them), and YOU are not looking in the log
for the errors (as suggested by Ashley and Mark).
2. You have a parse error online 2 as hinted by Mark: no semicolon to
terminate the line. You don't see this error due to #1
3. You have another parse error on line 4 between the double quote and
the variable you're trying to concatenate. You're missing the
concatenation operator: the 'dot' (.). You don't see this error due to #1
So, fix #1 either by turning on display_errors or read your logfiles.
Then you'll be able to identify problems #2 and #3 for yourself, and fix
them easily.
- Tul
---
This email has been checked for viruses by Avast antivirus software.
http://www.avast.com
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
